If a circular loop of wire of mass 'm' and radius 'R' is making 'n' revolutions per second about a point on its rim perpendicular to the plane of the loop, Then its rotational kinetic energy will be

Options

A. `pi^(2)mR^(2)n^(2)`
B. `2pi^(2)mR^(2)n^(2)`
C. `4pi^(2)mR^(2)n^(2)`
D. `8 pi^(2)mR^(2)n^(2)`

Explanation

To find the rotational kinetic energy of a circular loop of wire of mass 'm' and radius 'R' making 'n' revolutions per second about a point on its rim, we can follow these steps:

### Step 1: Understand the concept of rotational kinetic energy
The rotational kinetic energy (K) of a rotating object is given by the formula:
\[ K = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

### Step 2: Determine the moment of inertia (I)
For a circular loop (which can be treated as a thin ring) of mass \( m \) and radius \( R \), the moment of inertia about an axis through its center is:
\[ I_{cm} = m R^2 \]

However, since the loop is rotating about a point on its rim (not through the center), we need to use the parallel axis theorem to find the moment of inertia about this new axis. The parallel axis theorem states:
\[ I = I_{cm} + m d^2 \]
where \( d \) is the distance from the center of mass to the new axis. In this case, \( d = R \) (the radius of the loop).

Thus, the moment of inertia about the point on the rim is:
\[ I = m R^2 + m R^2 = 2 m R^2 \]

### Step 3: Calculate the angular velocity (ω)
The angular velocity \( \omega \) can be calculated from the number of revolutions per second \( n \):
\[ \omega = 2 \pi n \]
This is because one complete revolution corresponds to an angle of \( 2\pi \) radians.

### Step 4: Substitute I and ω into the kinetic energy formula
Now we can substitute \( I \) and \( \omega \) into the rotational kinetic energy formula:
\[ K = \frac{1}{2} I \omega^2 \]
Substituting the values:
\[ K = \frac{1}{2} (2 m R^2) (2 \pi n)^2 \]

### Step 5: Simplify the expression
Now simplify the expression:
\[ K = \frac{1}{2} \cdot 2 m R^2 \cdot 4 \pi^2 n^2 \]
\[ K = 4 m R^2 \pi^2 n^2 \]

### Final Answer
Thus, the rotational kinetic energy of the circular loop of wire is:
\[ K = 4 m R^2 \pi^2 n^2 \]

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